Adapted by Peter Au, George Brown College McGraw-Hill Ryerson Copyright © 2011 McGraw-Hill Ryerson Limited. 8.1 8.2 8.3 8.4 8.5 8.6 Null and Alternative Hypotheses and Errors in Testing Type I and Type II Errors and Their Probabilities Z Tests about a Population Mean (s Known): One-Sided Alternatives Z Tests about a Population Mean (s Known): Two-Sided Alternatives t Tests about a Population Mean (s Unknown) Z Tests about a Population Proportion Copyright © 2011 McGraw-Hill Ryerson Limited 8-2 L01 • The null hypothesis, denoted H0, is a statement of the basic proposition being tested. The statement generally represents the status quo and is not rejected unless there is convincing sample evidence that it is false • The alternative or research hypothesis, denoted Ha, is an alternative (to the null hypothesis) statement that will be accepted only if there is convincing sample evidence that it is true Copyright © 2011 McGraw-Hill Ryerson Limited 8-3 L01 • With a new billing system, the mean bill paying time m is hoped • to be less than 19.5 days • The alternative hypothesis Ha is that the new billing system has a mean payment time that is less than 19.5 days • With the old billing system, the mean bill paying time m was close to but not less than 39 days, so much more than 19.5 days • The null hypothesis H0 is that the new billing system has a mean payment time close to but not less than 39 days (which is more than 19.5 days) • One-Sided, “Less Than” Alternative • H0: m 19.5 vs.Ha: m < 19.5 Copyright © 2011 McGraw-Hill Ryerson Limited 8-4 L01 • An important quality characteristic is the case hardness depth, in millimeters (mm), of the engine camshaft (read case 8.2) • The optimal target value is 4.5 mm • Take m = 4.5 mm • Test whether the case hardness is 4.5 mm on average (the null) or not (the alternative) • Two-Sided, “Not Equal To” Alternative • H0: m = 4.5 vs. Ha: m 4.5 Copyright © 2011 McGraw-Hill Ryerson Limited 8-5 L01 • One-Sided, “Greater Than” Alternative H0: m m0 vs. Ha: m > m0 • One-Sided, “Less Than” Alternative H0 : m m0 vs. Ha : m < m0 • Two-Sided, “Not Equal To” Alternative H0 : m = m0 vs. Ha : m m0 where m0 is a given constant value (with the appropriate units) that is a comparative value Copyright © 2011 McGraw-Hill Ryerson Limited 8-6 L01 • As a result of testing H0 vs. Ha, will have to decide either of the following decisions for the null hypothesis H0: • Do not reject H0 • A weaker statement than “accepting H0” • But you are rejecting the alternative Ha OR • Reject H0 • A weaker statement than “accepting Ha” Copyright © 2011 McGraw-Hill Ryerson Limited 8-7 L01 • In order to “test” H0 vs. Ha, use the “test statistic” x m0 x m0 z sx s n where m0 is the given value or the hypothesized mean (often the claimed to be true) and x-bar is the mean of a sample • If the population is normal, then the test statistic z follows a normal distribution but the parent population DOES NOT HAVE TO BE NORMAL Copyright © 2011 McGraw-Hill Ryerson Limited 8-8 L01 • In the payment time case, with m0 = 19.5, use the test statistic: x 19.5 x 19.5 z sx s n • If z ≤ 0, then x ≤ m0 and thus, there is evidence to support rejecting H0 in favor of Ha • The point estimate x indicates that m is probably less than 19.5 • If z > 0, then x > m0 and thus, there is no evidence to support rejecting H0 • The point estimate x indicates that m is probably greater than 19.5 • The larger the value of z (the farther x is above m), the stronger the evidence to support H0 Copyright © 2011 McGraw-Hill Ryerson Limited 8-9 L03 • Type I Error: Rejecting H0 when it is true • Type II Error: Failing to reject H0 when it is false State of Nature Conclusion H0 True H0 False Reject H0 Type I Error Correct Decision Do not Reject H0 Correct Decision Type II Error Copyright © 2011 McGraw-Hill Ryerson Limited 8-10 L03 • Type I Error: Rejecting H0 when it is true • α is the probability of making a Type I error • 1 – a is the probability of not making a Type I error • Type II Error: Failing to reject H0 when it is false • β is the probability of making a Type II error • 1 – b is the probability of not making a Type II error State of Nature Conclusion Reject H0 Do not Reject H0 Copyright © 2011 McGraw-Hill Ryerson Limited H0 True H0 False α 1-α 1-β β 8-11 L03 • Usually set a to a low value • So that there is only a small chance of rejecting a true H0 • Typically, a = 0.05 • For a = 0.05, strong evidence is required to reject H0 • Usually choose a between 0.01 and 0.05 • a = 0.01 requires very strong evidence is to reject H0 • Sometimes choose a as high as 0.10 • Tradeoff between a and b • For fixed sample size, the lower we set a, the higher is b • The higher a, the lower b Copyright © 2011 McGraw-Hill Ryerson Limited 8-12 L04 • Test hypotheses about a population mean using the normal distribution • These tests are called Z tests • They require that the true value of the population standard deviation s is known • In most real-world situations, s is not known • But often is estimated from s of a single sample • When s is unknown, test hypotheses about a population mean using the t distribution (later) • Here, assume that we know s • Test using a “rejection point rule” Copyright © 2011 McGraw-Hill Ryerson Limited 8-13 L04 The steps are as follows: 1. State the null and alternative hypotheses 2. Specify the significance level a 3. Select the test statistic 4. Determine the rejection rule for deciding whether or not to reject H0 5. Collect the sample data and calculate the value of the test statistic 6. Decide whether to reject H0 by using the test statistic and the rejection rule 7. Interpret the statistical results in managerial terms and assess their practical importance Copyright © 2011 McGraw-Hill Ryerson Limited 8-14 L04 • Cable TV companies claim that their full cable packages cost on average $50 per month • A marketing company wants to demonstrate that it is significantly higher 1. State the null and alternative hypotheses • H0: m 50 Ha: m > 50 • μ is the hypothesized mean • We modify the hypothesis to H0: m = 50 Ha: m > 50 • since if there is evidence to reject H0 at μ = 50 then there is evidence to reject it at μ ≤ 50 2. Specify the significance level a • a = 0.05 Copyright © 2011 McGraw-Hill Ryerson Limited 8-15 L04 3. Select the test statistic • Use the test statistic: z x 50 x 50 sx s n • A positive value of this this test statistic results from a sample mean that is greater than $50 • Which provides evidence against H0 in favor of Ha Copyright © 2011 McGraw-Hill Ryerson Limited 8-16 L04 4. Determine the rejection rule for deciding whether or not to reject H0 • To decide how large the test statistic must be to reject H0 by setting the probability of a Type I error to a, do the following: • The probability a is the area in the right-hand tail of the standard normal curve (one tailed test) • Use the normal table to find the point za (called the rejection or critical point) • za is the point under the standard normal curve that gives a right-hand tail area equal to a • Since a = 0.05 in the cable case, the rejection point is za = z0.05 = 1.645 Copyright © 2011 McGraw-Hill Ryerson Limited 8-17 L04 4. Continued • Reject H0 in favor of Ha if the test statistic z is greater than the rejection point za • This is the rejection rule • In the cable subscriptions example, the rejection rule is to reject H0 if the calculated test statistic z is > 1.645 Copyright © 2011 McGraw-Hill Ryerson Limited 8-18 L04 5. Collect the sample data and calculate the value of the test statistic • Assume that s is known and s = $1.65 • For a sample of n = 40, we observed a value of x $50.575 x 50 50.575 50 z 2.20 s n 1.65 40 Copyright © 2011 McGraw-Hill Ryerson Limited 8-19 L05 L06 6. Decide whether to reject H0 by using the test statistic and the rejection rule • Compare the value of the test statistic to the rejection point according to the rejection rule • In the cable case, z = 2.20 (test statistic) z0.05 = 1.645 (rejection point) 2.20 > 1.645 • Therefore reject H0: m ≤ 50 in favor of Ha: m > 50 at the 0.05 significance level • Have rejected H0 by using a test that allows only a 5% chance of wrongly rejecting H0 • This result is “statistically significant” at the 0.05 level Copyright © 2011 McGraw-Hill Ryerson Limited 8-20 L05 L06 7. Interpret the statistical results in managerial terms and assess their practical importance • We conclude that it appears as though the mean full cable subscription package is more than $50 Copyright © 2011 McGraw-Hill Ryerson Limited 8-21 L07 • Testing H0: m = 50 versus Ha: m > 50 for (a) α = 0.05 and (b) α = 0.01 Copyright © 2011 McGraw-Hill Ryerson Limited 8-22 L07 • At a = 0.01, the rejection point is z0.01 = 2.33 • In the previous example, the test statistic z = 2.20 is < z0.01 = 2.33 • Therefore, cannot reject H0 in favor of Ha at the a = 0.01 significance level • This is the opposite conclusion reached with a = 0.05 • So, the smaller we set a, the larger is the rejection point, and the stronger is the statistical evidence that is required to reject the null hypothesis H0 Copyright © 2011 McGraw-Hill Ryerson Limited 8-23 L08 • The p-value or the observed level of significance is the probability of the obtaining the sample result if the null hypothesis H0 is true • The p-value is a conditional probability • It measures the level of support there is for H0 given the observed data • If the p-value is low, then that suggests that your data does not support H0 • If the p-value is not low, then that suggests that your data does support H0 • Rules: • 1. If the p-value < a, then we MUST REJECT H0 • 2. If the p-value > a, then we CANNOT REJECT H0 • Note: The p-value is the smallest value of a for which we can reject H0 Copyright © 2011 McGraw-Hill Ryerson Limited 8-24 L08 • For a greater than alternative, the p-value is the probability of observing a value of the test statistic greater than or equal to z when H0 is true • Use p-value as an alternative to testing a greater than alternative with a z test statistic (rejection point) Copyright © 2011 McGraw-Hill Ryerson Limited 8-25 L08 • Testing H0: m 50 vs. Ha: m > 50 using rejection points in (a) and p-value in (b) • Have to modify steps 4, 5, and 6 of the previous procedure to use p-values Copyright © 2011 McGraw-Hill Ryerson Limited 8-26 L09 4. Collect the sample data and compute the value of the test statistic • In the cable subscriptions example, the value of the test statistic was calculated to be z = 2.20 5. Calculate the p-value by using the test statistic value • The area under the standard normal curve in the right-hand tail to the right of the test statistic z = 2.20 is the p-value • The area is 0.5 – 0.4861 = 0.0139 • The p-value is 0.0139 Copyright © 2011 McGraw-Hill Ryerson Limited 8-27 L09 L10 5. Continued • If H0 is true, the probability is 0.0139 of obtaining a sample whose mean is $50.575 or higher • This is a very low probability and it appears as though H0 is not true and should therefore be rejected 6. Rejection Rule: We will reject H0 if the p-value is less than a • In the cable case, a was set to 0.05 • The calculated p-value of 0.0139 is < a = 0.05 • This implies that the test statistic z = 2.20 is > the rejection point z0.05 = 1.645 • Therefore reject H0 at the a = 0.05 significance level Copyright © 2011 McGraw-Hill Ryerson Limited 8-28 The steps are as follows: 1. State the null and alternative hypotheses 2. Specify the significance level a 3. Select the test statistic 4. Determine the rejection rule for deciding whether or not to reject H0 5. Collect the sample data and calculate the value of the test statistic 6. Decide whether to reject H0 by using the test statistic and the rejection rule 7. Interpret the statistical results in managerial terms and assess their practical importance Copyright © 2011 McGraw-Hill Ryerson Limited 8-29 L10 1. State the null and alternative hypotheses • In the payment time case • H0: m ≥ 19.5 vs. Ha: m < 19.5, where m is the mean bill payment time (in days) 2. Specify the significance level a • In the payment time case, set a = 0.01 Copyright © 2011 McGraw-Hill Ryerson Limited 8-30 L10 3. Select the test statistic • In the payment time case, use the test statistic z x 19.5 sx x 19.5 s n • A negative value of this this test statistic results from a sample mean that is less than 19.5 days • Which provides evidence against H0 in favor of Ha Copyright © 2011 McGraw-Hill Ryerson Limited 8-31 L10 4. Determine the rejection rule for deciding whether or not to reject H0 • To decide how much less than 0 the test statistic must be to reject H0 by setting the probability of a Type I error to a, do the following: • The probability a is the area in the left-hand tail of the standard normal curve • Use normal table to find the rejection point –za • –za is the negative of za • –za is the point on the horizontal axis under the standard normal curve that gives a left-hand tail area equal to a Copyright © 2011 McGraw-Hill Ryerson Limited 8-32 L10 4. Continued • Because a = 0.01 in the payment time case, the rejection point is –za = –z0.01 = –2.33 • Reject H0 in favor of Ha if the test statistic z is calculated to be less than the rejection point ( –zα) • This is the rejection rule • In the payment time case, the rejection rule is to reject H0 if the calculated test statistic is less than –2.33 Copyright © 2011 McGraw-Hill Ryerson Limited 8-33 L10 5. Collect the sample data and calculate the value of the test statistic • In the payment time case, assume that s is known and s = 4.2 days • For a sample of n = 65, x-bar = 18.1077 days: x 19.5 18.1077 19.5 z 2.67 s n 4.2 65 Copyright © 2011 McGraw-Hill Ryerson Limited 8-34 L10 6. Decide whether to reject H0 by using the test statistic and the rejection rule • Compare the value of the test statistic to the rejection point according to the rejection rule • In the payment time case, z = –2.67 is less than z0.01 = –2.33 • Therefore reject H0: m ≥ 19.5 in favor of Ha: m < 19.5 at the 0.01 significance level • Have rejected H0 by using a test that allows only a 1% chance of wrongly rejecting H0 • This result is “statistically significant” at the 0.01 level Copyright © 2011 McGraw-Hill Ryerson Limited 8-35 L10 7. Interpret the statistical results in managerial terms and assess their practical importance • We can conclude that the mean bill payment time of the new billing system appears to be less than 19.5 days • This is a substantial improvement over the old system by more than 50% Copyright © 2011 McGraw-Hill Ryerson Limited 8-36 L09 • Testing H0: m ≥ 19.5 vs. Ha: m < 19.5 using rejection points in (a) and p-value in (b) • Have to modify steps 4, 5, and 6 of the previous procedure to use pvalues Copyright © 2011 McGraw-Hill Ryerson Limited 8-37 L09 4. Collect the sample data and compute the value of the test statistic • In the payment time case, the value of the test statistic was calculated to be z = –2.67 5. Calculate the p-value using the test statistic value • In the payment time case, the area under the standard normal curve in the left-hand tail to the left of the test statistic z = –2.67 • The area is 0.5 – 0.4962 = 0.0038 • The p-value is 0.0038 Copyright © 2011 McGraw-Hill Ryerson Limited 8-38 L10 5. Continued • If H0 is true, then the probability is 0.0038 of obtaining a sample whose mean is as low as 18.1077 days or lower • This is a very low probability and it appears as though H0 is not true and should therefore be rejected 6. Reject H0 if the p-value is less than a • In the payment time case, a was 0.01 • The calculated p-value of 0.0038 is < a = 0.01 • This implies that the test statistic z = –2.67 is less than the rejection point –z0.01 = –2.33 • Therefore, reject H0 at the a = 0.01 significance level Copyright © 2011 McGraw-Hill Ryerson Limited 8-39 • If s is known we can use a z test to test H0 • We can reject H0: m = m0 at the a level of significance (probability of Type I error equal to a) if and only if the appropriate rejection point rule holds Test Statistic z x m0 s n Copyright © 2011 McGraw-Hill Ryerson Limited Alternative Reject H0 if: Ha : m > m0 z > -za Ha : m < m0 z < -za 8-40 • If the population variance is known, we can reject H0: m = m0 at the a level of significance (probability of Type I error equal to a) if and only if the corresponding p-value is less than the specified a Test Statistic z x m0 s n Copyright © 2011 McGraw-Hill Ryerson Limited Alternative Reject H0 if: Ha : m > m0 p value < a Ha : m < m0 p value < a 8-41 • Testing a “not equal to” alternative hypothesis • The sampling distribution of all possible sample means is modeled by a normal curve • Require that the true value of the population standard deviation s is known • What’s new and different here? • The area corresponding to the significance level a is evenly split into both the right- and left-hand tails of the standard normal curve Copyright © 2011 McGraw-Hill Ryerson Limited 8-42 The steps are as follows: 1. State the null and alternative hypotheses 2. Specify the significance level a 3. Select the test statistic 4. Determine the rejection rule for deciding whether or not to reject H0 5. Collect the sample data and calculate the value of the test statistic 6. Decide whether to reject H0 by using the test statistic and the rejection rule 7. Interpret the statistical results in managerial terms and assess their practical importance Copyright © 2011 McGraw-Hill Ryerson Limited 8-43 L06 1. State the null and alternative hypotheses • In the camshaft case, H0: m = 4.5 vs. Ha: m ≠ 4.5, where m is the mean hardness depth of the camshaft (in mm) 2. Specify the significance level a • In the camshaft case, set a = 0.05 3. Select the test statistic • In the camshaft case, use the test statistic: z Copyright © 2011 McGraw-Hill Ryerson Limited x 4.5 sx x 4.5 s n 8-44 L06 3. Continued • A positive value of this test statistic results from a sample mean that is greater than 4.5 mm • Which provides evidence against H0 and for Ha • A negative value of this this test statistic results from a sample mean that is less than 4.5 mm • Which also provides evidence against H0 and for Ha • A very small value close to 0 (either very slightly positive or very slightly negative) of this this test statistic results from a sample mean that is nearly 4.5 mm • Which provides evidence in favor of H0 and against Ha Copyright © 2011 McGraw-Hill Ryerson Limited 8-45 L06 4. Determine the rejection rule for deciding whether or not to reject H0 • To decide how different the test statistic must be from zero (positive or negative) to reject H0 in favor of Ha by setting the probability of a Type I error to a, do the following: • Divide the value of a in half to get a/2. This tail probability will be distributed in each tail of the standard normal curve (a two-tailed test), with a/2 in the left tail and a/2 in the right tail Copyright © 2011 McGraw-Hill Ryerson Limited 8-46 L06 4. Continued • Use the normal table to find the rejection points za/2 and –za/2 • za/2 is the point on the horizontal axis under the standard normal curve that gives a right-hand tail area equal to a/2 • –za/2 is the point on the horizontal axis under the standard normal curve that gives a left-hand tail area equal to a/2 Copyright © 2011 McGraw-Hill Ryerson Limited 8-47 L04 4. Continued • Because a = 0.05 in the camshaft case, a/2=0.025 • The area under the standard normal to the right of the mean is 0.5 – 0.025 = 0.475 • From Table A.3, the area is 0.475 for z = 1.96 • The rejection points are za = z0.025 = 1.96 OR –za = – z0.025 = – 1.96 • Reject H0 in favour of Ha if the test statistic z satisfies either: 1. z greater than the rejection point za/2 or 2. –z less than the rejection point –za/2 » This is the rejection rule Copyright © 2011 McGraw-Hill Ryerson Limited 8-48 L04 Copyright © 2011 McGraw-Hill Ryerson Limited 8-49 L04 5. Collect the sample data and calculate the value of the test statistic • In the camshaft case, assume that s is known and s = 0.47 mm • For a sample of n = 40, x-bar = 4.26 mm • Then z x 4.5 s n Copyright © 2011 McGraw-Hill Ryerson Limited 4.26 4.5 0.47 3.02 35 8-50 L06 6. Decide whether to reject H0 by using the test statistic and the rejection rule • Compare the value of the test statistic to the rejection point according to the rejection rule • In the camshaft case, z = –3.02 is less than – z0.025 = –1.96 • Therefore we must reject H0: m = 4.5 in favor of Ha: m ≠ 4.5 at the 0.05 significance level • We have rejected H0 using a test that allows for only a 5% chance of wrongly rejecting H0 • This result is said to be “statistically significant” at the 5% level of significance Copyright © 2011 McGraw-Hill Ryerson Limited 8-51 L06 7. Interpret the statistical results in managerial terms and assess their practical importance • We can conclude that it appears as though the mean hardness depth of the camshaft is not 4.5 mm • Using x = 4.26 and s = 0.47, we estimate that 99.73 percent of all individual camshaft hardness depths are in the interval [ x 3s ] 4.26 30.47) 2.85,5.67 • This estimated tolerance interval says that, because of a somewhat small mean of 4.26, we estimate that we are producing some hardness depths that are below the lower specification limit of 3 mm • Changes must be made to the manufacturing process Copyright © 2011 McGraw-Hill Ryerson Limited 8-52 L04 • Testing H0: m = 4.5 versus Ha: m ≠ 4.5 for a = 0.05 Copyright © 2011 McGraw-Hill Ryerson Limited 8-53 L09 4. Collect the sample data and compute the value of the test statistic • In the camshaft case, the value of the test statistic was calculated to be z = –3.02 5. Calculate the one-sided p-value using the test statistic value • In the camshaft case, the area under the standard normal curve in the left-hand tail to the left of the test statistic value z = –3.02 • The area is 0.5 – 0.4987 = 0.0013 • The one-sided p-value is 0.0013 Copyright © 2011 McGraw-Hill Ryerson Limited 8-54 L09 5. Continued • We are conducting a two-sided test though, and therefore, need a two-sided p-value. All we need to do is double the one-sided pvalue • Our two-sided p-value would be 2(0.0013) = 0.0026 • This is a very low probability and it appears as though H0 is not true and should therefore be rejected 6. Reject H0 if the two-sided p-value is less than a • In the camshaft case, a was 0.05 • The calculated two-sided p-value of 0.0026 is < a = 0.05 • Therefore reject H0 at the a = 0.05 significance level • We would also reject H0 at the 1% significance level Copyright © 2011 McGraw-Hill Ryerson Limited 8-55 L09 • Using a z statistic to test H0 when the population variance is known, we can reject H0: m = m0 at the a level of significance (probability of Type I error equal to a) if and only if: Test Statistic Alternative Reject H0 if: z > za/2 x m0 z s n Ha : m ≠ m0 (1) or 2-sided p-value < a z < –za/2 (2) or 2-sided p-value < a 1. 2. a/2 is the area under the standard normal curve to the right of za/2 a/2 is the area under the standard normal curve to the left of –za/2 Copyright © 2011 McGraw-Hill Ryerson Limited 8-56 L09 • Calculate the test statistic and the corresponding p-value • You can rate the strength of the conclusion, in general, about H0 according to these guidelines: • With a p-value < 0.10, there is some evidence to reject H0 • With a p-value < 0.05, there is strong evidence to reject H0 • With a p-value < 0.01, there is very strong evidence to reject H0 • With a p-value < 0.001, there is extremely strong evidence to reject H0 • So, the smaller the p-value, the stronger the evidence is against Ho Copyright © 2011 McGraw-Hill Ryerson Limited 8-57 • The null hypothesis H0 can be rejected in favor of the alternative Ha at some level of significance, a, if and only if the 100(1 – a)% confidence interval for m does not contain m0 • where m0 is the claimed value for the population mean • In other words, if the confidence interval does not contain the claimed value m0, then the null hypothesis can be rejected • Conversely, if the confidence interval does contain the claimed value m0, then the null hypothesis cannot be rejected Copyright © 2011 McGraw-Hill Ryerson Limited 8-58 •In the camshaft case, the 95% confidence interval for m is s 0.47 x za 2 4.26 1.96 4.10 , 4.42 n 35 •which does not contain target value of 4.5 mm •Therefore we must reject the null hypothesis that m is 4.5 mm at the 0.05 significance level •Here, the confidence level 1 – a = 0.95 and the significance level a = 0.05 •These sum to 1, so here see that the confidence level and significance levels are complementary Copyright © 2011 McGraw-Hill Ryerson Limited 8-59 • The a we use here is the same a that we used in Chapter 7 • The confidence level is 1 – a • The significance level is a • (1 – a ) + a = 1, so the confidence level and significance levels are complementary Copyright © 2011 McGraw-Hill Ryerson Limited 8-60 • Suppose the population being sampled is normally distributed • The population standard deviation s is unknown, as is the usual situation • If the population standard deviation s is unknown, then it will have to estimated from a sample standard deviation s • Under these two conditions, have to use the t distribution to test hypotheses • The t distribution from Chapter 7 Copyright © 2011 McGraw-Hill Ryerson Limited 8-61 • Define a new random variable t: t x m s n • with the definition of symbols as before • The sampling distribution of this random variable is a t distribution with n – 1 degrees of freedom Copyright © 2011 McGraw-Hill Ryerson Limited 8-62 • Let x-bar be the mean of a sample of size n with standard deviation s • Also, m0 is the claimed value of the population mean • Define a new test statistic t x m0 s n • If the population being sampled is normal, and • If s is used to estimate s, then … • The sampling distribution of the t statistic is a t distribution with n – 1 degrees of freedom Copyright © 2011 McGraw-Hill Ryerson Limited 8-63 • Same rejection rules apply here that applied with our Z-tests • Reject H0: m = m0 in favor of a particular alternative hypothesis Ha at the a level of significance if and only if the test statistic lies in the appropriate rejection region or, equivalently, if the corresponding p-value is less than a • We have the following rules … Copyright © 2011 McGraw-Hill Ryerson Limited 8-64 L09 Alternative Reject H0 if: p-value (reject Ho if p-value < a) Ha: m > m0 t > ta Area under t distribution to right of t Ha: m < m0 t < –ta Area under t distribution to left of –t Ha: m m0 |t| > t a/2 * Twice area under t distribution to right of |t| • ta, ta/2, and p-values are based on n – 1 degrees of freedom (for a sample of size n) • * either t > ta/2 or t < –ta/2 Copyright © 2011 McGraw-Hill Ryerson Limited 8-65 • In 2001, the mean credit card interest rate is as it was at the rate of 18.8% • A financial officer wishes to study whether the increased competition in the credit card business has reduced interest rates • This is the null hypothesis H0: m = 18.8 • The alternative to be tested is that the current mean interest rate isn’t as it was back in 2001, but in fact has decreased • This is the alternative hypothesis Ha: m < 18.8 • Test at the a = 0.05 level of significance • If H0 can be rejected at the 5% level, then conclude that it appears as though the mean rate now is less than 18.8% rate charged in 2001 Copyright © 2011 McGraw-Hill Ryerson Limited 8-66 L04 • We randomly select n = 15 credit cards • We know: n = 15 x = 16.828 s = 1.538 a = 0.05 • Then df = n – 1 = 14 • We will assume the population of the rates of all cards is approximately normal Copyright © 2011 McGraw-Hill Ryerson Limited 8-67 L04 • The rejection rule is: Reject H0: m = 18.8 against Ha: m < 18.8 if t < ta • For a = 0.05 and with df = 14, ta = t0.05 =1.761 • Calculate the value of the t statistic: t x 18.8 s n 16.827 18.8 1.538 15 4.97 • Because t = 4.97 < t0.05 = 1.761, reject H0 • Therefore, conclude at 5% significance level that it appears as though the mean interest rate is lower than it was in 2001 • In fact it was observed to be 18.8 – 16.827 = 1.973% lower • Our observed result of 16.827 is said to be “statistically significant” at a = 5% Copyright © 2011 McGraw-Hill Ryerson Limited 8-68 L09 L10 • The p-value here is the left-hand tail area under the t curve with df = 14 to the left of t = 4.97 • Our observed t value is off the t table, so using the table, we would say that our p-value < 0.0005. • A statistical software package can easily calculate the exact pvalue. It turns out that the p-value is actually 0.000103, which is less than 0.0005. • This says that if the claimed mean of 18.8 % is true, then there is only about a 1/10,000 chance of randomly selecting a sample of 15 credit cards whose mean rate would be as low as or lower than 16.827% • This p-value is < 0.05, 0.01, 0.001 • We would reject H0 at the 0.05, 0.01, 0.001 significance levels • Our result is highly significant Copyright © 2011 McGraw-Hill Ryerson Limited 8-69 L09 Copyright © 2011 McGraw-Hill Ryerson Limited 8-70 • If the sample size n is large, we can reject H0: p = p0 at the a level of significance (probability of Type I error equal to a) if and only if the appropriate rejection point condition holds or, equivalently, if the corresponding p-value is less than a • We have the following rules … Copyright © 2011 McGraw-Hill Ryerson Limited 8-71 Alternative Reject H0 if: p-value Ha: p > p0 z > zα Area under standard normal to the right of z Ha: p < p0 z < -zα Area under standard normal to the left of –z Ha: p ≠ p0 |z| > zα/2 * Twice the area under standard normal to the right of |z| • The test statistic is: z p̂ p0 p0 1 p0 ) n * either z > za/2 or z < –za/2 Copyright © 2011 McGraw-Hill Ryerson Limited 8-72 L06 • Virol has been shown to provide relief for 70% of all patients suffering from upper respiratory infections. A drug company believes that its competing drug, Phantol, is more effective than Virol • Let’s test Ho at a = 5% using rejection points and pvalues. • Test H0: p = 0.70 versus Ha: p > 0.70 Copyright © 2011 McGraw-Hill Ryerson Limited 8-73 L06 • A random sample of 300 patients was taken and 231 indicated that the drug had effectively treated their upper respiratory infections 231 0.77 • So, we have n = 300 and p 300 ^ • In order to test Ho, we need a test statistic: z ˆp p 0 p 0 1 p 0 ) n Copyright © 2011 McGraw-Hill Ryerson Limited 0.77 - 0.70 0.70 1 0.70) 300 2.65 8-74 L06 • Reject H0: p = 0.70 if the value of Z is greater than zA = z0.05 = 1.645 • Note that using this procedure is valid because np0 = 300(0.70) = 210 and n(1 - p0) = 300(1 - 0.70) = 90 are both at least 5 • Comparing values: • z = 2.65 > z0.05 = 1.645 > z0.01 = 2.33 < z0.001 = 3.09 • So reject H0 at the 0.05 and 0.01 significance levels but do not reject at the 0.001 level Copyright © 2011 McGraw-Hill Ryerson Limited 8-75 L09 • For z = 2.65, the p-value is: p-value = P(z > 2.65) = (0.5 – 0.4960) = 0.004 (table A.3) • But p = 004 < a = 0.05 < a = 0.01 > a = 0.001 • So (again) reject H0 at the 0.05 and 0.01 significance levels but do not reject at the 0.001 level Copyright © 2011 McGraw-Hill Ryerson Limited 8-76 • Hypothesis tests are used to test the strength of a statement for a population • The null hypothesis H0 is the statement being tested, it is the status quo, it is not rejected for the alternative hypothesis Ha unless there is convincing statistical evidence otherwise • There are three common ways to test a hypothesis 1) rejection region 2) p-value 3) confidence intervals • P-values can give more information as it is the value of α at which the decision will change • Types of parameters which we test in this chapter are means, proportions • Type I errors occur when a true null hypothesis is rejected and type II errors occur when a false null hypothesis is not rejected Copyright © 2011 McGraw-Hill Ryerson Limited 8-77 Copyright © 2011 McGraw-Hill Ryerson Limited 8-78